Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include "bits/stdc++.h"

using namespace std;

typedef long long ll;
typedef pair<int,int> P;
const double EPS = 1e-12;
const int INF = numeric_limits<int>::max()/2;
const int MOD = 1e9+7;

typedef vector<vector<int>> Graph;

vector<int> dijkstra(const Graph &g, int st){
    vector<int> d(g.size(),INF);
    d[st] = 0;
    priority_queue<P, vector<P>, greater<P>> que;
    que.push(P(0,st)); // P(dis, pos)
    while(!que.empty()){
        int dist = que.top().first;
        int v = que.top().second;
        que.pop();
        if(d[v]<dist) continue;
        for(const auto &to: g[v]){
            if(d[to]<=d[v]+1) continue;
            d[to] = d[v] + 1;
            que.push(P(d[to], to));
        }
    }
    return d;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n,m,st,gt;cin>>n>>m>>st>>gt;
    st--;gt--;
    Graph g(n);
    for(int i=0;i<m;i++){
        int x,y; cin>>x>>y; x--;y--;
        g[x].push_back(y);
        g[y].push_back(x);
    }
    vector<int> ds=dijkstra(g,st),dt=dijkstra(g,gt);
    int dis=ds[gt];
    if(dis==1){
        cout<<0<<endl;
    }
    else{
        map<int,int> mp;
        for(int i=0;i<n;i++){
            int d=dt[i];
            mp[d]++;
        }
        ll res=0;
        for(int i=0;i<n;i++){
            int d=dis-2-ds[i];
            res+=mp[d];
        }
        cout<<res<<endl;
    }
}